# Sum of all the prime divisors of a number

• Difficulty Level : Medium
• Last Updated : 23 May, 2022

Given a number N. The task is to find the sum of all the prime divisors of N.

Examples:

```Input: 60
Output: 10
2, 3, 5 are prime divisors of 60

Input: 39
Output: 16
3, 13 are prime divisors of 39```

A naive approach will be to iterate for all numbers till N and check if the number divides N. If the number divides N, check if that number is prime or not. Add all the prime numbers till N which divides N.

Below is the implementation of the above approach:

## C++

 `// CPP program to find sum of``// prime divisors of N``#include ``using` `namespace` `std;``#define N 1000005` `// Function to check if the``// number is prime or not.``bool` `isPrime(``int` `n)``{``    ``// Corner cases``    ``if` `(n <= 1)``        ``return` `false``;``    ``if` `(n <= 3)``        ``return` `true``;` `    ``// This is checked so that we can skip``    ``// middle five numbers in below loop``    ``if` `(n % 2 == 0 || n % 3 == 0)``        ``return` `false``;` `    ``for` `(``int` `i = 5; i * i <= n; i = i + 6)``        ``if` `(n % i == 0 || n % (i + 2) == 0)``            ``return` `false``;` `    ``return` `true``;``}` `// function to find sum of prime``// divisors of N``int` `SumOfPrimeDivisors(``int` `n)``{``    ``int` `sum = 0;``    ``for` `(``int` `i = 1; i <= n; i++) {``        ``if` `(n % i == 0) {``            ``if` `(isPrime(i))``                ``sum += i;``        ``}``    ``}``    ``return` `sum;``}``// Driver code``int` `main()``{``    ``int` `n = 60;``    ``cout << ``"Sum of prime divisors of 60 is "` `<< SumOfPrimeDivisors(n) << endl;``}`

## C

 `// C program to find sum of``// prime divisors of N``#include ``#include ` `#define N 1000005` `// Function to check if the``// number is prime or not.``bool` `isPrime(``int` `n)``{``    ``// Corner cases``    ``if` `(n <= 1)``        ``return` `false``;``    ``if` `(n <= 3)``        ``return` `true``;` `    ``// This is checked so that we can skip``    ``// middle five numbers in below loop``    ``if` `(n % 2 == 0 || n % 3 == 0)``        ``return` `false``;` `    ``for` `(``int` `i = 5; i * i <= n; i = i + 6)``        ``if` `(n % i == 0 || n % (i + 2) == 0)``            ``return` `false``;` `    ``return` `true``;``}` `// function to find sum of prime``// divisors of N``int` `SumOfPrimeDivisors(``int` `n)``{``    ``int` `sum = 0;``    ``for` `(``int` `i = 1; i <= n; i++) {``        ``if` `(n % i == 0) {``            ``if` `(isPrime(i))``                ``sum += i;``        ``}``    ``}``    ``return` `sum;``}` `// Driver code``int` `main()``{``    ``int` `n = 60;``    ``printf``(``"Sum of prime divisors of 60 is %d\n"``,SumOfPrimeDivisors(n));``}` `// This code is contributed by kothavvsaakash.`

## Java

 `// Java program to find sum``// of prime divisors of N``import` `java.io.*;``import` `java.util.*;` `class` `GFG``{``// Function to check if the``// number is prime or not.``static` `boolean` `isPrime(``int` `n)``{``    ``// Corner cases``    ``if` `(n <= ``1``)``        ``return` `false``;``    ``if` `(n <= ``3``)``        ``return` `true``;` `    ``// This is checked so that``    ``// we can skip middle five``    ``// numbers in below loop``    ``if` `(n % ``2` `== ``0` `|| n % ``3` `== ``0``)``        ``return` `false``;` `    ``for` `(``int` `i = ``5``;``             ``i * i <= n; i = i + ``6``)``        ``if` `(n % i == ``0` `||``            ``n % (i + ``2``) == ``0``)``            ``return` `false``;` `    ``return` `true``;``}` `// function to find``// sum of prime``// divisors of N``static` `int` `SumOfPrimeDivisors(``int` `n)``{``    ``int` `sum = ``0``;``    ``for` `(``int` `i = ``1``;``             ``i <= n; i++)``    ``{``        ``if` `(n % i == ``0``)``        ``{``            ``if` `(isPrime(i))``                ``sum += i;``        ``}``    ``}``    ``return` `sum;``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``int` `n = ``60``;``    ``System.out.print(``"Sum of prime divisors of 60 is "` `+``                          ``SumOfPrimeDivisors(n) + ``"\n"``);``}``}`

## C#

 `// C# program to find sum``// of prime divisors of N``using` `System;``class` `GFG``{``    ` `// Function to check if the``// number is prime or not.``static` `bool` `isPrime(``int` `n)``{``    ``// Corner cases``    ``if` `(n <= 1)``        ``return` `false``;``    ``if` `(n <= 3)``        ``return` `true``;` `    ``// This is checked so that``    ``// we can skip middle five``    ``// numbers in below loop``    ``if` `(n % 2 == 0 || n % 3 == 0)``        ``return` `false``;` `    ``for` `(``int` `i = 5;``             ``i * i <= n; i = i + 6)``        ``if` `(n % i == 0 ||``            ``n % (i + 2) == 0)``            ``return` `false``;` `    ``return` `true``;``}` `// function to find``// sum of prime``// divisors of N``static` `int` `SumOfPrimeDivisors(``int` `n)``{``    ``int` `sum = 0;``    ``for` `(``int` `i = 1;``            ``i <= n; i++)``    ``{``        ``if` `(n % i == 0)``        ``{``            ``if` `(isPrime(i))``                ``sum += i;``        ``}``    ``}``    ``return` `sum;``}` `// Driver code``public` `static` `void` `Main()``{``    ``int` `n = 60;``    ``Console.WriteLine(``"Sum of prime divisors of 60 is "` `+``                        ``SumOfPrimeDivisors(n) + ``"\n"``);``}``}` `// This code is contributed``// by inder_verma.`

## Python3

 `# Python 3 program to find``# sum of prime divisors of N``N ``=` `1000005` `# Function to check if the``# number is prime or not.``def` `isPrime(n):``    ` `    ``# Corner cases``    ``if` `n <``=` `1``:``        ``return` `False``    ``if` `n <``=` `3``:``        ``return` `True` `    ``# This is checked so that ``    ``# we can skip middle five``    ``# numbers in below loop``    ``if` `n ``%` `2` `=``=` `0` `or` `n ``%` `3` `=``=` `0``:``        ``return` `False` `    ``i ``=` `5``    ``while` `i ``*` `i <``=` `n:``        ``if` `(n ``%` `i ``=``=` `0` `or``            ``n ``%` `(i ``+` `2``) ``=``=` `0``):``            ``return` `False``        ``i ``=` `i ``+` `6` `    ``return` `True` `# function to find sum``# of prime divisors of N``def` `SumOfPrimeDivisors(n):``    ``sum` `=` `0``    ``for` `i ``in` `range``(``1``, n ``+` `1``) :``        ``if` `n ``%` `i ``=``=` `0` `:``            ``if` `isPrime(i):``                ``sum` `+``=` `i``    ` `    ``return` `sum` `# Driver code``n ``=` `60``print``(``"Sum of prime divisors of 60 is "` `+``              ``str``(SumOfPrimeDivisors(n)))` `# This code is contributed``# by ChitraNayal`

## PHP

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## Javascript

 ``

Output

`Sum of prime divisors of 60 is 10`

Time Complexity: O(N * sqrt(N))

Efficient Approach: The complexity can be reduced using Sieve of Eratosthenes with some modifications. The modifications are as follows:

• Take an array of size N and substitute zero in all the indexes(initially consider all the numbers are prime).
• Iterate for all the numbers whose indexes have zero(i.e., it is prime numbers).
• Add this number to all it’s multiples less than N
• Return the array[N] value which has the sum stored in it.

Below is the implementation of the above approach.

## C++

 `// CPP program to find prime divisors of``// all numbers from 1 to n``#include ``using` `namespace` `std;` `// function to find prime divisors of``// all numbers from 1 to n``int` `Sum(``int` `N)``{``    ``int` `SumOfPrimeDivisors[N+1] = { 0 };` `    ``for` `(``int` `i = 2; i <= N; ++i) {` `        ``// if the number is prime``        ``if` `(!SumOfPrimeDivisors[i]) {` `            ``// add this prime to all it's multiples``            ``for` `(``int` `j = i; j <= N; j += i) {` `                ``SumOfPrimeDivisors[j] += i;``            ``}``        ``}``    ``}``    ``return` `SumOfPrimeDivisors[N];``}` `// Driver code``int` `main()``{``    ``int` `N = 60;``    ``cout << ``"Sum of prime divisors of 60 is "``         ``<< Sum(N) << endl;``}`

## Java

 `// Java program to find``// prime divisors of``// all numbers from 1 to n``import` `java.io.*;``import` `java.util.*;` `class` `GFG``{``    ` `// function to find prime``// divisors of all numbers``// from 1 to n``static` `int` `Sum(``int` `N)``{``    ``int` `SumOfPrimeDivisors[] = ``new` `int``[N + ``1``];``    `  `    ``for` `(``int` `i = ``2``; i <= N; ++i)``    ``{` `        ``// if the number is prime``        ``if` `(SumOfPrimeDivisors[i] == ``0``)``        ``{` `            ``// add this prime to``            ``// all it's multiples``            ``for` `(``int` `j = i; j <= N; j += i)``            ``{` `                ``SumOfPrimeDivisors[j] += i;``            ``}``        ``}``    ``}``    ``return` `SumOfPrimeDivisors[N];``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``int` `N = ``60``;``    ``System.out.print(``"Sum of prime "` `+``                ``"divisors of 60 is "` `+``                       ``Sum(N) + ``"\n"``);``}``}`

## Python3

 `# Python 3 program to find``# prime divisors of``# all numbers from 1 to n` `# function to find prime``# divisors of all numbers``# from 1 to n``def` `Sum``(N):`` ` `    ``SumOfPrimeDivisors ``=` `[``0``] ``*` `(N ``+` `1``)``     ` `    ``for` `i ``in` `range``(``2``, N ``+` `1``) :``     ` `        ``# if the number is prime``        ``if` `(SumOfPrimeDivisors[i] ``=``=` `0``) :``         ` `            ``# add this prime to``            ``# all it's multiples``            ``for` `j ``in` `range``(i, N ``+` `1``, i) :``             ` `                ``SumOfPrimeDivisors[j] ``+``=` `i``             ` `    ``return` `SumOfPrimeDivisors[N]`` ` `# Driver code``N ``=` `60``print``(``"Sum of prime"` `,``      ``"divisors of 60 is"``,``                  ``Sum``(N));``                  ` `# This code is contributed``# by Smitha`

## C#

 `// C# program to find``// prime divisors of``// all numbers from 1 to n``using` `System;` `class` `GFG``{``    ` `// function to find prime``// divisors of all numbers``// from 1 to n``static` `int` `Sum(``int` `N)``{``    ``int` `[]SumOfPrimeDivisors = ``new` `int``[N + 1];``    ` `    ``for` `(``int` `i = 2; i <= N; ++i)``    ``{` `        ``// if the number is prime``        ``if` `(SumOfPrimeDivisors[i] == 0)``        ``{` `            ``// add this prime to``            ``// all it's multiples``            ``for` `(``int` `j = i;``                     ``j <= N; j += i)``            ``{` `                ``SumOfPrimeDivisors[j] += i;``            ``}``        ``}``    ``}``    ` `    ``return` `SumOfPrimeDivisors[N];``}` `// Driver code``public` `static` `void` `Main()``{``    ``int` `N = 60;``    ``Console.Write(``"Sum of prime "` `+``                    ``"divisors of 60 is "` `+``                         ``Sum(N) + ``"\n"``);``}``}` `// This code is contributed``// by Smitha`

## PHP

 ``

## Javascript

 ``

Output

`Sum of prime divisors of 60 is 10`

Time Complexity: O(N * log N)

#### Efficient Approach:

Time complexity can be reduced by finding all the factors efficiently.

Below approach describe how to find all the factors efficiently.

If we look carefully, all the divisors are present in pairs. For example if n = 100, then the various pairs of divisors are: (1,100), (2,50), (4,25), (5,20), (10,10)

Using this fact we could speed up our program significantly.

We, however, have to be careful if there are two equal divisors as in the case of (10, 10). In such case, we’d take only one of them.

Below is the implementation of the above approach.

## C++

 `// C++ program to find sum of``// prime divisors of N``#include ``using` `namespace` `std;` `// Function to check if the``// number is prime or not.``bool` `isPrime(``int` `n)``{``    ``// Corner cases``    ``if` `(n <= 1)``        ``return` `false``;``    ``if` `(n <= 3)``        ``return` `true``;` `    ``// This is checked so that we can skip``    ``// middle five numbers in below loop``    ``if` `(n % 2 == 0 || n % 3 == 0)``        ``return` `false``;` `    ``for` `(``int` `i = 5; i * i <= n; i = i + 6)``        ``if` `(n % i == 0 || n % (i + 2) == 0)``            ``return` `false``;` `    ``return` `true``;``}` `// function to find sum of prime``// divisors of N``int` `SumOfPrimeDivisors(``int` `n)``{``    ``int` `sum = 0;``    ``// return type of sqrt function``    ``// if float``    ``int` `root_n = (``int``)``sqrt``(n);``    ``for` `(``int` `i = 1; i <= root_n; i++) {``        ``if` `(n % i == 0) {``            ``// both factors are same``            ``if` `(i == n / i && isPrime(i)) {``                ``sum += i;``            ``}``            ``else` `{``                ``// both factors are``                ``// not same ( i and n/i )``                ``if` `(isPrime(i)) {``                    ``sum += i;``                ``}``                ``if` `(isPrime(n / i)) {``                    ``sum += (n / i);``                ``}``            ``}``        ``}``    ``}``    ``return` `sum;``}``// Driver code``int` `main()``{``    ``int` `n = 60;``    ``cout << ``"Sum of prime divisors of 60 is "``         ``<< SumOfPrimeDivisors(n) << endl;``}``// This code is contributed by hemantraj712`

## C

 `// C program to find sum of``// prime divisors of N``#include ``#include ``#include ` `// Function to check if the``// number is prime or not.``bool` `isPrime(``int` `n)``{``    ``// Corner cases``    ``if` `(n <= 1)``        ``return` `false``;``    ``if` `(n <= 3)``        ``return` `true``;` `    ``// This is checked so that we can skip``    ``// middle five numbers in below loop``    ``if` `(n % 2 == 0 || n % 3 == 0)``        ``return` `false``;` `    ``for` `(``int` `i = 5; i * i <= n; i = i + 6)``        ``if` `(n % i == 0 || n % (i + 2) == 0)``            ``return` `false``;` `    ``return` `true``;``}` `// function to find sum of prime``// divisors of N``int` `SumOfPrimeDivisors(``int` `n)``{``    ``int` `sum = 0;``    ``// return type of sqrt function``    ``// if float``    ``int` `root_n = (``int``)``sqrt``(n);``    ``for` `(``int` `i = 1; i <= root_n; i++) {``        ``if` `(n % i == 0) {``            ``// both factors are same``            ``if` `(i == n / i && isPrime(i)) {``                ``sum += i;``            ``}``            ``else` `{``                ``// both factors are``                ``// not same ( i and n/i )``                ``if` `(isPrime(i)) {``                    ``sum += i;``                ``}``                ``if` `(isPrime(n / i)) {``                    ``sum += (n / i);``                ``}``            ``}``        ``}``    ``}``    ``return` `sum;``}``// Driver code``int` `main()``{``    ``int` `n = 60;``    ``printf``(``"Sum of prime divisors of 60 is %d\n"``,SumOfPrimeDivisors(n));``}``// This code is contributed by hemantraj712`

## Java

 `// Java program to find sum of``// prime divisors of N``class` `GFG{` `// Function to check if the``// number is prime or not.``static` `boolean` `isPrime(``int` `n)``{``    ` `    ``// Corner cases``    ``if` `(n <= ``1``)``        ``return` `false``;``    ``if` `(n <= ``3``)``        ``return` `true``;``  ` `    ``// This is checked so that we can skip``    ``// middle five numbers in below loop``    ``if` `(n % ``2` `== ``0` `|| n % ``3` `== ``0``)``        ``return` `false``;``  ` `    ``for``(``int` `i = ``5``; i * i <= n; i = i + ``6``)``        ``if` `(n % i == ``0` `|| n % (i + ``2``) == ``0``)``            ``return` `false``;``  ` `    ``return` `true``;``}``  ` `// Function to find sum of prime``// divisors of N``static` `int` `SumOfPrimeDivisors(``int` `n)``{``    ``int` `sum = ``0``;``    ` `    ``// Return type of sqrt function``    ``// if float``    ``int` `root_n = (``int``)Math.sqrt(n);``    ``for``(``int` `i = ``1``; i <= root_n; i++)``    ``{``        ``if` `(n % i == ``0``)``        ``{``            ` `            ``// Both factors are same``            ``if` `(i == n / i && isPrime(i))``            ``{``                ``sum += i;``            ``}``            ``else``            ``{``                ` `                ``// Both factors are``                ``// not same ( i and n/i )``                ``if` `(isPrime(i))``                ``{``                    ``sum += i;``                ``}``                ``if` `(isPrime(n / i))``                ``{``                    ``sum += (n / i);``                ``}``            ``}``        ``}``    ``}``    ``return` `sum;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `n = ``60``;``    ``System.out.println(``"Sum of prime divisors of 60 is "` `+``                       ``SumOfPrimeDivisors(n));``}``}` `// This code is contributed by divyeshrabadiya07`

## Python3

 `# Python3 program to find sum of``# prime divisors of N``import` `math` `# Function to check if the``# number is prime or not.``def` `isPrime(n) :` `    ``# Corner cases``    ``if` `(n <``=` `1``) :``        ``return` `False``    ``if` `(n <``=` `3``) :``        ``return` `True`` ` `    ``# This is checked so that we can skip``    ``# middle five numbers in below loop``    ``if` `(n ``%` `2` `=``=` `0` `or` `n ``%` `3` `=``=` `0``) :``        ``return` `False``    ` `    ``i ``=` `5``    ``while` `i ``*` `i <``=` `n :``        ``if` `(n ``%` `i ``=``=` `0` `or` `n ``%` `(i ``+` `2``) ``=``=` `0``) :``            ``return` `False``        ``i ``=` `i ``+` `6`` ` `    ``return` `True`` ` `# function to find sum of prime``# divisors of N``def` `SumOfPrimeDivisors(n) :` `    ``Sum` `=` `0``    ` `    ``# return type of sqrt function``    ``# if float``    ``root_n ``=` `(``int``)(math.sqrt(n))``    ``for` `i ``in` `range``(``1``, root_n ``+` `1``) :``        ``if` `(n ``%` `i ``=``=` `0``) :``          ` `            ``# both factors are same``            ``if` `(i ``=``=` `(``int``)(n ``/` `i) ``and` `isPrime(i)) :``                ``Sum` `+``=` `i``            ` `            ``else` `:``                ``# both factors are``                ``# not same ( i and n/i )``                ``if` `(isPrime(i)) :``                    ``Sum` `+``=` `i``                ` `                ``if` `(isPrime((``int``)(n ``/` `i))) :``                    ``Sum` `+``=` `(``int``)(n ``/` `i)``                   ` `    ``return` `Sum``    ` `n ``=` `60``print``(``"Sum of prime divisors of 60 is"``, SumOfPrimeDivisors(n))` `# This code is contributed by rameshtravel07`

## C#

 `// C# program to find sum of``// prime divisors of N``using` `System;``class` `GFG {``    ` `    ``// Function to check if the``    ``// number is prime or not.``    ``static` `bool` `isPrime(``int` `n)``    ``{``        ``// Corner cases``        ``if` `(n <= 1)``            ``return` `false``;``        ``if` `(n <= 3)``            ``return` `true``;``     ` `        ``// This is checked so that we can skip``        ``// middle five numbers in below loop``        ``if` `(n % 2 == 0 || n % 3 == 0)``            ``return` `false``;``     ` `        ``for` `(``int` `i = 5; i * i <= n; i = i + 6)``            ``if` `(n % i == 0 || n % (i + 2) == 0)``                ``return` `false``;``     ` `        ``return` `true``;``    ``}``     ` `    ``// function to find sum of prime``    ``// divisors of N``    ``static` `int` `SumOfPrimeDivisors(``int` `n)``    ``{``        ``int` `sum = 0;``        ``// return type of sqrt function``        ``// if float``        ``int` `root_n = (``int``)Math.Sqrt(n);``        ``for` `(``int` `i = 1; i <= root_n; i++) {``            ``if` `(n % i == 0) {``                ``// both factors are same``                ``if` `(i == n / i && isPrime(i)) {``                    ``sum += i;``                ``}``                ``else` `{``                    ``// both factors are``                    ``// not same ( i and n/i )``                    ``if` `(isPrime(i)) {``                        ``sum += i;``                    ``}``                    ``if` `(isPrime(n / i)) {``                        ``sum += (n / i);``                    ``}``                ``}``            ``}``        ``}``        ``return` `sum;``    ``}` `  ``static` `void` `Main() {``    ``int` `n = 60;``    ``Console.WriteLine(``"Sum of prime divisors of 60 is "` `+ SumOfPrimeDivisors(n));``  ``}``}` `// This code is contributed by suresh07.`

## Javascript

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Output

`Sum of prime divisors of 60 is 10`

Time Complexity: O(sqrt(N) * sqrt(N))

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